Question

please answer this....​

Answer 1

Let the given equation be P(n).

P(n) : 3³ⁿ - 26n - 1

For P(1) : 3³ - 26 - 1 = 27 - 27 = 0

Hence, the given equation is divisible by 676.

Let for any k ∈ N, given expression is divisible by 676.

Then, $33^{k}$ − 26k − 1 = 676c ___(1), where c is a natural number.

Now, we have to prove, for n = k + 1 , given expression is divisible by 676 if it is true for k.

For n = k + 1 , given expression is,  

$3^{3(k+1)}-26(k+1)-1\\ =27.3^{3k}-26k-26-1$

Using (1) equation;

= 27 ( 676c + 26k + 1 ) − 26k − 27

= 27 ( 676c ) + 27 ( 26k ) + 27 − 26k − 27

= 27 ( 676c ) + 27 ( 26k ) − 26k

= 27 ( 676c ) + 26 ( 26k )

= 27 ( 676c ) + 676k , which is divisible by 676.

Thus, given expression will be divisible by 676.

Answer 2

Let the given equation be P(n).

P(n) : 3³ⁿ - 26n - 1

For P(1) : 3³ - 26 - 1 = 27 - 27 = 0

Hence, the given equation is divisible by 676.

Let for any k ∈ N, given expression is divisible by 676.

Then, 33^{k}33

k

− 26k − 1 = 676c ___(1), where c is a natural number.

Now, we have to prove, for n = k + 1 , given expression is divisible by 676 if it is true for k.

For n = k + 1 , given expression is,

\begin{lgathered}3^{3(k+1)}-26(k+1)-1\\ =27.3^{3k}-26k-26-1\end{lgathered}

3

3(k+1)

−26(k+1)−1

=27.3

3k

−26k−26−1

Using (1) equation;

= 27 ( 676c + 26k + 1 ) − 26k − 27

= 27 ( 676c ) + 27 ( 26k ) + 27 − 26k − 27

= 27 ( 676c ) + 27 ( 26k ) − 26k

= 27 ( 676c ) + 26 ( 26k )

= 27 ( 676c ) + 676k , which is divisible by 676.

Thus, given expression will be divisible by 676.